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Physics
UTME / JAMB 1985
Electricity and Magnetism
Solution
A wire P has half the diameter and half the length of a wire Q of similar material. The ratio of the resistance of P to that of Q is
A
8 : 1
B
4 : 1
C
2 : 1
D
1 : 1
E
1 : 4
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Resistance \(\alpha\) \(\frac{length}{area}\)
\(\frac{R_Q}{R_P}\) = \(\frac{L_QA_P}{L_PA_Q}\)
LQ = 2LP
DQ = 2DP
AP = \(\pi\)\(\frac{(2D_P)^2}{2}\)
= \(\pi\)DP
AQ = \(\pi\)\(\frac{(D_P)^2}{2}\)
= \(\frac{1}{4}\)\(\pi\)DP
therefore, \(\frac{R_Q}{R_P}\) = \(\frac{2L_P \times \pi D_P}{L_P \times \frac{1}{4} \pi D_P}\)
= 2 : 1
Source: Myschool Past Questions Archive (Record #28648)
Physics
UTME / JAMB 1985
Electricity and Magnetism
Solution
A transformer has a primary coil with 500 turns and a secondary coil with 2500 turns. When the voltage input to the primary coil is 120V, the output is
A
6000V
B
600V
C
240V
D
60V
E
24V
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
For transformers, \(\frac{V_p}{V_s}\) = \(\frac{N_p}{N_s}\) = \(\frac{I_p}{I_s}\)
\(\frac{500}{2500}\) = \(\frac{120}{V_s}\)
Vs = 600v
Source: Myschool Past Questions Archive (Record #28655)
Physics
UTME / JAMB 1984
Electricity and Magnetism
The unit of quantity of electricity is called
A
the ampere
B
the volts
C
the coulomb
D
the ammeter
E
electromotive force
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #28676)
Physics
UTME / JAMB 1984
Electricity and Magnetism
The resistance of a wire depends on
A
the length of the wire
B
the diameter of the wire
C
the temperature of the wire
D
the resistivity of the wire
E
all of the above
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option E.
Source: Myschool Past Questions Archive (Record #28677)
Physics
UTME / JAMB 1984
Electricity and Magnetism
Solution
Two cells each of e.m.f. 1.5V and an internal resistance 2\(\Omega\) are connected in parallel. Calculate the current flowing when the cells are connected to a 1\(\Omega\) resistor
A
0.75\(\Omega\)
B
1.5\(\Omega\)
C
0.5\(\Omega\)
D
1.0\(\Omega\)
E
0.6\(\Omega\)
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
E = 1.5v
rT = \(\frac{2 \times 2}{2 + 2}\)
= 1
e.m.f. = I(R + r)
1.5 = I(1 + 1)
I = 0.75\(\Omega\)
Source: Myschool Past Questions Archive (Record #28679)
Physics
UTME / JAMB 1984
Electricity and Magnetism
Diagram
Solution
What is the resistance of the circuit shown above
Attached Figure / Diagram
A
4Ω
B
11Ω
C
19/5Ω
D
17/4Ω
E
8Ω
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
from the diagram, 2ohms and 2ohms are parallel, then total resistance = 1ohms
1ohms, 3ohms and the 4ohms are in series
then total resistance = 1 + 3 + 4 = 8ohms.
Source: Myschool Past Questions Archive (Record #41642)
Physics
UTME / JAMB 1984
Electricity and Magnetism
Diagram
Solution
A short bar magnet is placed in earth's magnetic field, as shown above, which of the points II and III are/is possible positions of a neutral point?
Attached Figure / Diagram
A
I only
B
II only
C
III only
D
I and II
E
I,II and III
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
When the magnet is placed with its north pole pointing toward the earth's magnetic north, the two neutral points will occur at III, according to the diagram above.
Source: Myschool Past Questions Archive (Record #41644)
Physics
UTME / JAMB 1983
Electricity and Magnetism
Solution
A milliameter with full scale deflection of 10mA has an internal resistance of 5 ohms. It would be converted to an ammeter with a full scale deflection of 1A by connecting a resistance of
A
\(\frac{5}{99}\)ohms in series with it
B
\(\frac{5}{99}\)ohms in parallel with it
C
\(\frac{99}{5}\)ohms in series with it
D
\(\frac{99}{5}\)ohms in parallel with it
E
2 ohms in series with it
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
I = 1 - (10 x 10-3)
= 0.99A
IR = 10 x 10 x 10-3 x 5
R = \(\frac{5}{99}\)\(\Omega\)
Source: Myschool Past Questions Archive (Record #28842)
Physics
UTME / JAMB 1983
Electricity and Magnetism
Solution
A 24V potential difference is applied across a parallel combination of four 6-ohms resistor, The current in each resistor is
A
1A
B
4A
C
16A
D
18A
E
36A
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
\(I = \frac{24}{6}\)
= 4A
Source: Myschool Past Questions Archive (Record #28851)
Physics
UTME / JAMB 1983
Electricity and Magnetism
Solution
The electrochemical equivalent of a metal is 0.126 x 10-6 kgC-1. The mass of the metal that a current of 5A will deposit from a suitable bath in 1 hour is
A
0.0378 x 10 -3kg
B
0.227 x 10 -3kg
C
0.378 x 10 -3kg
D
0.595 x 10 -3kg
E
2.268 x 10 -3kg
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
m = Itz
= 5 x 3600 x 0.126 x 10-6
= 2.268 x 10-3kg
Source: Myschool Past Questions Archive (Record #28855)
Physics
UTME / JAMB 1983
Electricity and Magnetism
Diagram
Solution
Four resistors R\(_1\),R\(_2\),R\(_3\), and R\(_4\) are connected in series as shown above, V\(_1\),V\(_2\),V\(_3\) are voltmeters connected as indicated above. Which of the following relations is correct??
Attached Figure / Diagram
A
V1 = V3 =V2/2
B
V1 = 2V2 = V3
C
V1=1/2V3 =V3
D
V2 - V1 = V3
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
From the diagram: The four resistors (\(R_1, R_2, R_3, R_4\)) are in series, so the same current (I) flows through all of them.
The voltmeters are connected as follows:
Voltmeter
Measures the voltage across
\(V_1\)
\(R_1\)
\(V_2\)
\(R_2 + R_3\)
\(V_3\)
\(R_4\)
Using Ohm’s Law:
V = IR
So, \(V_1 = IR_1\)
\(V_2 = I(R_2 + R_3)\)
\(V_3 = IR_4\)
In this type of question, the resistors are usually assumed equal:
\(R_1 = R_2 = R_3 = R_4\) = R
Calculate the voltages
\(V_1\) = IR
\(V_2\) = I(R + R) = 2IR
\(V_3\) = IR
Compare them
\(V_1 = V_3\)
and \(V_2 = 2V_1\)
]Therefore: \(V_1 = V_3 = \frac{V_2}{2}\)
Source: Myschool Past Questions Archive (Record #41637)
Physics
UTME / JAMB 1983
Electricity and Magnetism
Diagram
In the Figure above, QR is a vertical conductor and the current I flows from R to Q, P is a point on the horizontal plane and is to the south of the wire. The direction of the magnetic field at P due to the current is
Attached Figure / Diagram
A
Upward
B
North
C
South
D
West
E
East
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #41638)
Physics
UTME / JAMB 1982
Electricity and Magnetism
Solution
An electric iron is rated at 1000watts, 250v. The corresponding maximum resistance and accompanying current is
A
62.5A, 4.0\(\Omega\)
B
16.0A, 62.5\(\Omega\)
C
62.5A, 16.0\(\Omega\)
D
4.0A, 62.5\(\Omega\)
E
4.0A, 250\(\Omega\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
R = \(\frac{V^2}{R}\) = IV ; R = \(\frac{V^2}{P}\) and I = \(\frac{P}{V}\)
R = \(\frac{(250)^2}{100}\) and I = R = \(\frac{1000}{250}\)
R = 62.5\(\Omega\) and I = 4A
Source: Myschool Past Questions Archive (Record #28896)
Physics
UTME / JAMB 1982
Electricity and Magnetism
As a positively charge rod is brought nearer to the top cap of a positively charged gold leaf electroscope, the divergence of the leaves will
A
decrease to zero
B
steadily increase
C
decrease to zero and then increase
D
remain constant
E
increase to a maximum and then decrease
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #28925)
Physics
UTME / JAMB 1982
Electricity and Magnetism
Solution
A high-resistance voltmeter reads 3.0V when connected across the terminals of a battery on open circuit and 2.6V when the battery supplies a current of 0.2A through a lamp. The resistance of lamp is
A
2.00\(\Omega\)
B
13.00\(\Omega\)
C
0.52\(\Omega\)
D
0.13\(\Omega\)
E
1.50\(\Omega\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
IR = 2.6 and Ir = 0.4
0.2 x R = 2.6
R = 13.00\(\Omega\)
Source: Myschool Past Questions Archive (Record #28930)